Maths Qstion (1 Viewer)

[Paj20]

A commemorative cricket ball has a diameter of 7cm. It is to be preserved in a cubic case that will allow 5mm on each side of the ball.

(A) What will the side length of the cubic case be = 8cm

(B) Caculate the amount of empty space in side the case, to the nearest whole number... the book says 300cm?? is that right

 

[Slidey]

Break the problem down.

We have a cube, so what is the side length? If the diameter of the ball is 7cm and we need 1mm freedom on each side of the ball, than we have 7cm + 2*1mm = 7.2cm.

The cube has side length 7.2cm. If you meant 1cm freedom each side, then it would be 9cm. 8cm is not correct. 8cm is correct if you accidentally forgot to type the 5 before mm.

Let us assume it is 5mm either side:

Onto the question:

What is a cube? A square in 3 dimensions, so if one side length is 8cm, then all of them are. So the volume is fairly simple: multiply the area of a square by the side length: (8*8)*8 = 2^3^3=2^9=512 cm cubed.

What is the volume of a circle? From memory it is (pi*diam^3)/6 = pi*7^3/6 cm cubed which I'll check on the calc in a sec.

Anyway, to find volume not occupied by ball, you could draw a Venn diagram or go: Cube Vol - Ball Vol ~= 332.41 cm cubed, nearest whole would be 332 cm cubed.

 

[the-derivative]

Hmmm well I get 332.41 - which is kind of close.
I got my answer like this:

Side of Cube = 8cm (given - I'm assuming its correct - but I'm not too sure)
Volume of Cube = 512
Volume of Sphere (ball) = 4/3 x pi x (3.5)^3 = 197.59cm
then Empty space = 512-197.59 = 332.41

I probably made a mistake somewhere - but thats what I got. Btw did u get 8cm from the textbook?

EDIT: Woops sorry - didnt see Slidey's post. You beat me too it.

 

[Mark576]

If you're having trouble with (a) consider this: It seems (from the answers) that they want us to construct visually a cricket ball that has been pushed to a corner inside a cube. Now from what I can infer, you mean to say cm, not mm, and that the closest distance from the sides of the cube not touching the ball to the ball (which has been pushed into a corner for simplification) is 1cm. From this it follows that the side length of the cubic case will be 1cm + 7cm = 8cm. For (B) the amount of empty space would be calculated as follows:

V (empty space) = V (cubic case) - V (cricket ball) = 8^3 - (4/3*pi*r^3) where r = 3.5cm

The amount of empty space comes out to be 332.41cm^3, which (to the nearest hundred) = 300cm^3

EDIT: Ah, beaten to it.

EDIT 2: In case anyone is wondering, my answer has been modelled to assume that the answer to (a) is 8cm. But I agree that Slidey's answer is more correct.

 

[Paj20]

lol yep i get it now thanks heaps

oh and the answer in the back of the book was 332

 

[aznboystar]

i got another one, how do u differentiate 6^x ?